A particle executes a linear S.H.M. of amplitude A. At what dis"tan"ce from the mean position is its K.E. equal to its P.E. ?

Options

A. 0.51 A
B. 0.61 A
C. 0.81 A
D. 0.71 A

Answer

D

Explanation

P.E. = K.E.
`therefore (1)/(2)Kx^(2)=(1)/(2)K(A^(2)-x^(2))=(1)/(2)KA^(2)-(1)/(2)Kx^(2)`
`therefore 2x^(2)=A^(2) therefore x=(A)/(sqrt(2))=(A)/(1.414)=0.71 A`